Question

Set up the integral to compute the volume of the solid that lies above the cone z =root(x^2+y^2) and below sphere x^2+y^2+z^2=4

Answer 1

z=r^2
r^2+z^2=4
z=sqrt(4-r^2)
use rdrd(thet)dz
\[\int\limits_{0}^{2\pi} \int\limits_{0}^{\sqrt{2}} \int\limits_{\sqrt{4-r^2}}^{r}dzrdrd \theta\]

Answer 2

the r and sqrt 4-r^2 shood be reversed sorry

Answer 3

Thank you very much man.

Answer 4

it might be right or i can do sperical coordinates

Answer 5

i think the cylindrical coordinates here work just fine.

thanks